The physics of a falling Slinky

What happens when you drop a Slinky?  The answer, in this video, may surprise you.

And when you attach a tennis ball to the bottom?

The Slinky, like Silly Putty, was an accident: the spring was originally intended to support naval instruments.  The Ur-Slinky was accidentally dropped, “walked” down several objects, and the rest is history.

h/t: Science Goddess and Ed Yong via Matthew Cobb

46 Comments

  1. Torbjorn Larsson, OM
    Posted September 27, 2011 at 3:42 am | Permalink

    The system behavior, where information has to travel, connects back to the ftl (or not) neutrinos of course.

    • Posted September 27, 2011 at 6:00 am | Permalink

      Speaking of which…have you had a chance to see if your idea of the discrepancy being explained by the difference between a straight-line measurement and one following the curvature of the Earth is correct?

      Cheers,

      b&

  2. MKray
    Posted September 27, 2011 at 3:51 am | Permalink

    I see the fall of the slinky as a superposition of two motions: (i) a fall of the centre of mass of the slinky, and, (ii) a compression of the slinky. It seems that the slinky that we see has the property that the sum of these motions leads to a stationary lower end. I can imagine a somewhat-slinky-like spring with different elastic properties for which that cancellation does not occur.

    • Posted September 27, 2011 at 4:03 am | Permalink

      I don’t think so ~ explain the version with the tennis ball attached at the bottom

      • Posted September 27, 2011 at 4:22 am | Permalink

        Consider… If an even heavier weight than the tennis ball was attached at the bottom & the spring was allowed to stretch to the greater length required to support the weight. And then the top was released ~ the weight would remain motionless until the upward pull lessened. The upward pull would remain sufficient to support the weight until the collapsed spring hits the weight.

        • MKray
          Posted September 27, 2011 at 4:33 am | Permalink

          Irrespective of any other feature, the centre of mass of the entire system will fall with the acceleration of gravity, apart from corrections for air resistance. The internal dynamics of the system will depend on the masses of the ball and the density and Hooke’s constant of slinky-stuff. There is nothing deeply mysterious about this… it just happens that the parameters are such as to end up leaving the end point (nearly) stationary.

          • Chris Booth
            Posted September 28, 2011 at 12:45 pm | Permalink

            Adding the tennis ball lowered the center of mass of the whole system from where it was when the slinky was un-coupled to a proportionally large mass.

      • Ken Pidcock
        Posted September 27, 2011 at 4:26 am | Permalink

        You mean the version where you can see the tennis ball falling a bit as the slinky comes down?

        • Drew
          Posted September 27, 2011 at 8:01 am | Permalink

          that slight downward motion of the ball is simply due to the fact that it wasn’t allowed to come to equilibrium before he let go of the slinky. If you watch it you see the ball/slinky is still bouncing slightly before he let go, and at the moment at which he does let go the ball/slinky is moving toward the ground. If he had let go of the slinky when the ball was in the upward movement of its bounce the tennis ball would have appeared to move up.

          Basically, while the compression wave, as he calls it, moves toward the bottom of the slinky, the bottom of the slinky and the attached ball continue doing whatever it was that they were doing when the front end is released until the wave makes contact with the bottom; in this case, they were slightly bouncing.

      • Posted September 27, 2011 at 6:45 pm | Permalink

        Whatever weight you hang from the slinky, the slinky has to become that more tense to hold it up, and has to lose that tension before the weight will start to fall.

    • Posted October 10, 2011 at 10:09 am | Permalink

      So now the Aussie geniuses have released a new video: Super-sized Slinky Drop This clearly demonstrates that the bottom of the Slinky…

      D o e s N o t M o v e

  3. Posted September 27, 2011 at 3:54 am | Permalink

    Lovely ~ my favourite product line of the manufacturers is Slinky Crazy Eyes ~ the spectacles with the plastic eyeballs on springs.

    The guy on the right in the video is the delightful Aussie tennis nut Prof. Rod Cross ~ The University of Sydney School of Physics. He retired in 2003 as an Honorary member of staff & continues to work on the physics of sport and forensic physics.

    Here’s his page full of interesting bits’n’pieces.

    Warning: Slinky has dangerous side effects ~ the inventor split from his wife, family & business to become a missionary in Bolivia

    • Diane G.
      Posted September 27, 2011 at 3:21 pm | Permalink

      A couple of interesting links, Michael–esp. the first. Thanks!

      • Posted September 27, 2011 at 3:49 pm | Permalink

        Thank you back Diane. It’s great to have people like him around ~ curiosity undimmed by age

  4. Kevin Alexander
    Posted September 27, 2011 at 4:22 am | Permalink

    I don’t get how this is counter intuitive. When I played with slinkies as a kid I noticed the same phenomenon. You don’t need a slowmo camera. If you stare at the bottom of the slinky when drop it you see that it doesn’t start to fall when you let go of the top. The top, on the other hand, falls faster (I didn’t use the word accelerate at the time) than a ball dropped.

    The spring is still stretched at the bottom when you let go of the top so, why would it start to fall? The top, on the other hand has both gravity plus the spring force so it goes quicker than some object let to fall. Adding a tennis ball doesn’t change any of that.

    I also don’t get why physicists need to resort to information theory when it isn’t necessary. A big reason why there is an anti science bigotry is that scientists themselves have a bad habit of overexplaining things.

    Tumsup’s Law “There is no phenomenon, however simple, where the explication cannot expand to the limits of the intellect of the person explaining it”

    Or as we used to say at work, “An engineer is someone who can describe a wheel in three hundred words without using the word ’round’ “

    • Brad
      Posted September 27, 2011 at 4:44 am | Permalink

      I like your point, Kevin, about overexplaining things. Very nice.

      • Kevin Alexander
        Posted September 27, 2011 at 5:23 am | Permalink

        My favourite example of Tumsup’s law

        How does an airplane fly? Almost every resource will show a diagram of a wing cross section and explain that the top of the wing is curved while the bottom is flat. the air will go faster over the top…Bernoulli…etc..etc.

        The problem with this explanation is that it’s not comprehensible to most people and, as anyone who has ever made a paper airplane could tell you, it isn’t necessary. If fact, most wings aren’t shaped that way, so why use it as an explanation?

        George Cayley described the real answer more than a century before the Wright Bros. If you incline a plane to the airflow then the air will strike one side and produce a force in the other direction. Perfectly understandable to anyone who’s ever stuck a flat hand out the car window. Why drag Bernoulli into it? Just to show off.

        • Michael Sternberg
          Posted September 27, 2011 at 8:11 am | Permalink

          Explaining the airfoil is not just to show off. After all, why go airplane manufacturers through the expense when a flat wing like in a paper plane would do? Taken farther than you perhaps intended, this kind of logic is rather anti-science and what we are accustomed to hear from Sarah Palin, say.

          Read about angle of attck.

          There is of course a challenging follow-up question connected here: Given all the business about Bernoulli etc., how is it that an air show plane can fly upside down?

          • Kevin Alexander
            Posted September 27, 2011 at 8:47 am | Permalink

            Michael,

            I acknowledge that It can be taken too far. My point was that when explaining something one should start with the simplest, most intuitive form. The aim should be to draw someone into science, not baffle them by going over their head with a more complex explanation.

            The curves of an airfoil are refinements of the basic plane, they are not the essential.

            It was George Cayley IIRC who coined the term angle of attack. He was certainly the first to study it experimentally.

            Sarah Palin probably thinks that the plane is held up by the prayers of the nervous passengers.

    • ChasCPeterson
      Posted September 27, 2011 at 5:15 am | Permalink

      Yes, excellent point I think.
      Dr. Coyne’s colleague and student H. Allen Orr says some complementary things about unnecessarily detailed overexplanation in his recent review of Brooks’s The Social Animal.

    • Filipe
      Posted September 27, 2011 at 8:14 am | Permalink

      You clearly don’t know what information theory is.

      The explanation given is the easiest way to understand what is happening. The disturbance needs time to travel.

      • Kevin Alexander
        Posted September 27, 2011 at 8:56 am | Permalink

        You’re right, I don’t understand it.

        You’re speaking a different language so maybe, for you, it is the easiest way to understand it.

        I do know slinkies though and I don’t think it takes a new language to explain them.

    • S.K.Graham
      Posted September 27, 2011 at 5:10 pm | Permalink

      Your point about tumsup’s law is spot on.

      And your example about airfoil/bernoulli over-explanation (actually, I would say non-explanation) is excellent, something that has bothered me for a long time as well. Flight is achieved by deflecting air downwards, thus producing the equal & opposite force upwards. If the force is greater than the weight of the object, it flies. That’s how birds do it. That’s how bees do it.

      However, I do see why the slinky is counter-intuitive. Your own “intuition” is the result of direct observation, so I’m not sure that counts as an intuition. Most people have the intuition that the hand is holding the slinky and once it lets go, the whole thing should fall. They are not thinking about the details of the internal tension forces of the slinky and how they change over time.

      The “explanation” provided certainly leaves something to be desired and tends to still make it sound “magical” (the information has to get from one end to the other before it “knows” to fall… say what?).

      The explanation is this: the stretched slinky is pulling up on itself with just enough force to cancel gravity (this is true at every point along the slinky). Immediately after you let go, the slinky is still stretched, so all parts of the slinky still have forces balanced, except the very top, which begins to fall. This reduces the “stretch” of the next bit of slinky down, which now has forces out of balance and can start to fall, and so on down the line.

  5. Hempenstein
    Posted September 27, 2011 at 10:17 am | Permalink

    I’d like to see a plain ball released at the exact same time as the slinky was released. Will they touch ground simultaneously, or will the Slinky get there first since some of the downward progression of the top is being pulled by the spring?

    Otherwise, until it was sold in the late 90’s, the company that makes the Slinky in PA only did that – they were more or less a large mom & pop operation. According to this, slinky is a Swedish word, and mom rescued the company from pop’s evangelically-motivated generosity: http://inventors.about.com/od/sstartinventions/a/slinky.htm

    • Posted September 27, 2011 at 10:34 am | Permalink

      That would be a great supplementary experiment to perform. A quick thought experiment shows that the solo ball will hit the ground first if the ground is close enough that the slinkified ball hasn’t started to move. I think that what you’d find for larger drops is that there’s a height that corresponds with the slinky’s center of mass such that the ball and the slinky will reach the ground at the same time.

      The really interesting bit would be comparing the acceleration profiles of the two systems….

      b&

      • Joe Clark
        Posted September 27, 2011 at 10:57 am | Permalink

        I would submit that the slinky would land first, if the ball was held at the same height as the top of the slinky.

        If the ball was held at the centre of mass of the slinky, I suspect that, discounting air resistance, the two should land at the same time.

        It seems to me that the centre of mass of the slinky should accelerate at the same speed as the ball, since the only other forces acting on the slinky are working in equal pairs either side of the centre of mass. (Not Newtonian pairs, mind you =P).

        • Posted September 27, 2011 at 11:41 am | Permalink

          Yep, that’s pretty much it.

          /@

        • Posted September 27, 2011 at 12:05 pm | Permalink

          Yes, you explained that better than I did.

          For the first experiment, I was envisioning (but didn’t state) that the balls would be at the same height when released, but the one would be suspended by the slinky. But find the center of mass of the slinky system, and the center of mass of both systems would fall at the same rate.

          The curious thing about the slinky system is that, as it stretches and compresses, the center of mass might well travel up and down the length of the coil, creating really counterintuitive visual effects. That’s why I’d love to see how it actually plays out.

          b&

        • Ichthyic
          Posted September 27, 2011 at 12:57 pm | Permalink

          I would submit that the slinky would land first, if the ball was held at the same height as the top of the slinky.

          don’t forget that the potential energy stored in the spring is added to the downward velocity when the top of the spring finally finishes releasing the full amount of potential energy as kinetic energy (well, and heat, of course).

          the ball has no additional potential energy to add to the equation.

          • S.K.Graham
            Posted September 27, 2011 at 5:16 pm | Permalink

            Actually Ichthyc, the potential energy of the stretch slinky is not relevant, because that energy will only be released internally causing the collapse followed by vibrations (if it were a normal spring which can be squeezed as well as stretched, then it would oscillate as it fell). It will not affect the rate at which the center of mass of the slinky falls.

            • Ichthyic
              Posted September 27, 2011 at 7:55 pm | Permalink

              actually, that would be correct if the slinky had enough time to fall for the oscillation to balance out.

              it does not.

              so, I’m right, and the slinky’s stored energy will indeed impart an initial faster downward velocity once the initial contraction is complete.

              you can even see it in the vid, as the slinky speeds up when the ends meet.

              • S.K.Graham
                Posted September 27, 2011 at 10:46 pm | Permalink

                You need to be careful in your terminology. If by “slinky” you mean the center of gravity of the slinky (or, equivalently, the average speed of the slinky as a whole), then, no, it does not “speed up”. When the “ends meet” the top part of the slinky suddenly slows down and the bottom suddenly speeds up (in a manner analogous to two objects that collide and stick together). If it appears to you that the slinky as a whole speeds up, it is just a visual illusion. I assure you, it does not. The average speed over all points of the slinky stays the same (or more precisely continues to accelerate at the rate of g, the acceleration of gravity). Think of it this way: the internal forces are preventing the lower parts of the slinky from accelerating at g, while simultaneously giving additional acceleration to the upper parts of the slinky, on average these effects cancel out, so the only thing contributing the the net momentum of the slinky is gravity.

                Internal forces cannot contribute to the acceleration of the center of gravity of an object or a system of interacting particles — because the momentum gained by any part of the system is exactly equal to momentum lost (motion in opposite direction resulting in “negative” momentum) by other parts of the system (action/reaction). “But”, you will say, “I see that the top gains momentum while the bottom sits still, neither losing nor gaining”. This is because the *external* force of gravity is exactly canceling the negative (upward) momentum that would be gained by the bottom of the spring if there was no gravity (think about what would happen if you could “turn off” gravity at the same moment you release the top of the spring).

                Neglecting air resistance, when the “top meets the bottom” the fully collapsed slinky has *exactly* the same momentum (and hence velocity) at the exact same time as it would if you released an identical fully collapsed slinky from the height of the slinky’s center of mass.

  6. Posted September 27, 2011 at 11:19 am | Permalink

    Wow. We always get these sorts of questions asked by the zanier breeds of physics teachers, but to see it in slow motion is quite impressive.

  7. Posted September 27, 2011 at 11:48 am | Permalink

    At release, the top of the slinky is experiencing spring tension downwards as well as gravitational force; so it accelerates faster than an object dropped from the same height. But the system center of mass is doing what it always does and hits the ground right on schedule.

    What needs explaining in this explanation is why the spring contracts only from one end rather than as a whole… clearly an explanation involving propagation delay is required, for which see about R/C electrical transmission lines.

    Next if we want to explain why the observed rates and intervals are what they are, we have to get into properties of the material and more details of initial conditions.

    Isn’t it wonderful how strikingly different forms of explanation bring out various aspects of this simple experiment? Whereas there’s the spring, just doing what it does all unknowing.

    • Joe Clark
      Posted September 27, 2011 at 12:00 pm | Permalink

      I believe the reason the spring appears to contract from the top, rather than uniformly, is that the top “links” are more extended than the bottom ones because of how the slinky hangs.
      It appears to me, although I’m happy to accept that I could be misinterpreting, that as the slinky becomes more compressed, the lower coils of the spring are visibly contracting at the same time as the upper ones.

      Does that sound reasonable?

      • Ichthyic
        Posted September 27, 2011 at 12:53 pm | Permalink

        yes. the reason the slinky does not contract from the bottom up on release is because of exactly what the guy says in the vid:

        the forces acting on the slinky are in equilibrium while he is holding it.

        the potential energy of the slinky is being maintained by his fingers holding it from the top, which is turned into kinetic energy as he releases the spring, BUT, it takes time for that energy to travel through the spring itself.

        so, the bottom of the spring does not change in equilibrium until the kinetic energy from the spring manages to travel to the end.

        energy does not travel instantaneously through objects.

    • Michael Sternberg
      Posted September 27, 2011 at 12:08 pm | Permalink

      Indeed, any “explanation” is tied to assumptions at many different levels. Here’s a classic piece of Feynman:

      • Posted September 27, 2011 at 1:44 pm | Permalink

        Thanks for that, very interesting. I wondered that the interviewer asked, and insisted on, the question “what is the feeling …” What sort of thing does he expect? I suppose he meant What is the feeling you get from magnetism? Maybe exactly what he wanted was something like, one way a rubber band pulling them together, the other way a spring pushing them apart. Richard thinks magnets are just as familiarly transparent as rubber bands, but maybe not.

        Original Richard is all over physics-type “why?” questions, naturally, and an excellent physics teacher he was. Instead of “Why is ice slippery?” someone might ask “Why did Auntie go standing on ice?” which leads off into the social/cultural situation and a different set of complications.

        So if you meet Richard in the street, avoid talking to him about feelings.

        • Posted September 27, 2011 at 1:49 pm | Permalink

          If you meet Richard in the street, blow his head off if you’re armed or run away if not, because all he’ll be interested in is your tasty, sweet, braaaaaaaaiiiiiiiinnnnnnnsssss.

          Cheers,

          b&

          • Posted September 27, 2011 at 3:14 pm | Permalink

            It’s true, Richard’s corporeal self has gone to the compost pile, but his brains linger among us: quantum electrodynamics is still considered useful, I believe.

            Besides which, my Dad was like that about feelings; he very much preferred explanations. He had an explicit case of Feynmann envy, as a matter of fact; Richard was a few classes ahead of him at school.

  8. Jim Thomerson
    Posted September 27, 2011 at 12:22 pm | Permalink

    On how airfoils work, to paraphrase; Bernoulli gets the air out of the way on top, and Newton pushes up from the bottom. At one time it was thought that a symmetrical airfoil cannot generate lift. True enough at 0 angle of incidence. I fly control line precision aerobatic airplanes with symmetrical airfoils. If you carefully watch one flying along level, you will see it is slightly nose up to get enough incidence to produce the requisite lift.

    • exrelayman
      Posted September 27, 2011 at 4:35 pm | Permalink

      To beat on this dead horse a little bit more, here is another perspective.

      An ultralight must be careful to avoid the air behind and below a conventional aircraft to avoid the ‘wake’ flow of air. This ‘wake’ air has been sent downward by the wings of the plane. The downward force exerted by the aircraft on the air is balanced by the upward force exerted by the air on the aircraft (enhanced by the Bernoulli effect), and when that force equals the force of gravity on the plane, level flight results.

      Seems reasonable to me. Is it?

      PS: Thanks Michael for the Feynman piece.

      • Posted September 27, 2011 at 5:09 pm | Permalink

        An ultralight must be careful to avoid the air behind and below a conventional aircraft to avoid the ‘wake’ flow of air.

        Ultralight, hell. Even a Learjet can get flipped by the wake from a 747.

        I distinctly remember driving up to the general aviation terminal at Albuquerque one time just as an airliner was landing. The ground just before the end of the runway was dusty dirt. You could very clearly see an impressive pair of counterrotating dust devils trailing the jet — it was the wingtip vortices aka wake turbulence. As we were driving there to leave in a Cessna 182 (a single-engine four-passenger propeller craft, the flying equivalent of the family sedan), it was a rather emphatic visual demonstration of just why you want to give the big boys lots of room.

        Air traffic controllers, of course, are well aware of such dangers and take great pains to not only keep aircraft separated by comfortable safety margins but to caution pilots whenever the theoretical possibility exists for the turbulence to last longer than reasonably expectable.

        And, yes, they caution 737 pilots landing behind 747 pilots — for good reason. The smaller aircraft can get tossed around like confetti, but even the big planes can experience sudden changes in attitude, and that’s not something you want to have happen when you’re 500 feet off the ground in any plane, let alone one with a 100-foot wingspan.

        Cheers,

        b&

  9. TrineBM
    Posted September 27, 2011 at 12:33 pm | Permalink

    Cool to see in slowmo!
    (And “Ur-Slinky” is now part of my active vocabulary)

  10. Ichthyic
    Posted September 27, 2011 at 12:46 pm | Permalink

    The answer, in this video, may surprise you.

    nope, sorry. I took freshman level physics.

    we learned about things like equilibrium, which is why also attaching a tennis ball to the bottom obviously wouldn’t have changed the outcome.

    though it was quite fun to see this demonstrated with a slow mo camera, rather that with equations on paper.
    :)


One Trackback/Pingback

  1. […] The falling slinky (hat tip: Jerry Coyne) […]

Follow

Get every new post delivered to your Inbox.

Join 30,611 other followers

%d bloggers like this: